Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
/**
 *    author:  tourist
 *    created: 25.01.2020 04:17:16       
**/
#include <bits/stdc++.h>

using namespace std;

template <typename A, typename B>
string to_string(pair<A, B> p);

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p);

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p);

string to_string(const string& s) {
  return '"' + s + '"';
}

string to_string(const char* s) {
  return to_string((string) s);
}

string to_string(bool b) {
  return (b ? "true" : "false");
}

string to_string(vector<bool> v) {
  bool first = true;
  string res = "{";
  for (int i = 0; i < static_cast<int>(v.size()); i++) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(v[i]);
  }
  res += "}";
  return res;
}

template <size_t N>
string to_string(bitset<N> v) {
  string res = "";
  for (size_t i = 0; i < N; i++) {
    res += static_cast<char>('0' + v[i]);
  }
  return res;
}

template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto &x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}

template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}

template <typename A, typename B, typename C>
string to_string(tuple<A, B, C> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ")";
}

template <typename A, typename B, typename C, typename D>
string to_string(tuple<A, B, C, D> p) {
  return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " + to_string(get<2>(p)) + ", " + to_string(get<3>(p)) + ")";
}

void debug_out() { cerr << endl; }

template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}

#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 42
#endif

class segtree {
 public:
  struct node {
    int val = -1;
    bool put = false;

    void apply(int l, int r, int v) {
      val = v;
      put = true;
    }
  };

  node unite(const node &a, const node &b) const {
    node res;
    return res;
  }

  inline void push(int x, int l, int r) {
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    if (tree[x].put) {
      tree[x + 1].apply(l, y, tree[x].val);
      tree[z].apply(y + 1, r, tree[x].val);
      tree[x].put = false;
    }
  }

  inline void pull(int x, int z) {
    tree[x] = unite(tree[x + 1], tree[z]);
  }

  int n;
  vector<node> tree;

  void build(int x, int l, int r) {
    if (l == r) {
      return;
    }
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    build(x + 1, l, y);
    build(z, y + 1, r);
    pull(x, z);
  }

  template <typename M>
  void build(int x, int l, int r, const vector<M> &v) {
    if (l == r) {
      tree[x].apply(l, r, v[l]);
      return;
    }
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    build(x + 1, l, y, v);
    build(z, y + 1, r, v);
    pull(x, z);
  }

  node get(int x, int l, int r, int ll, int rr) {
    if (ll <= l && r <= rr) {
      return tree[x];
    }
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    push(x, l, r);
    node res{};
    if (rr <= y) {
      res = get(x + 1, l, y, ll, rr);
    } else {
      if (ll > y) {
        res = get(z, y + 1, r, ll, rr);
      } else {
        res = unite(get(x + 1, l, y, ll, rr), get(z, y + 1, r, ll, rr));
      }
    }
    pull(x, z);
    return res;
  }

  template <typename... M>
  void modify(int x, int l, int r, int ll, int rr, const M&... v) {
    if (ll <= l && r <= rr) {
      tree[x].apply(l, r, v...);
      return;
    }
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    push(x, l, r);
    if (ll <= y) {
      modify(x + 1, l, y, ll, rr, v...);
    }
    if (rr > y) {
      modify(z, y + 1, r, ll, rr, v...);
    }
    pull(x, z);
  }

  int find_first_knowingly(int x, int l, int r, const function<bool(const node&)> &f) {
    if (l == r) {
      return l;
    }
    push(x, l, r);
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    int res;
    if (f(tree[x + 1])) {
      res = find_first_knowingly(x + 1, l, y, f);
    } else {
      res = find_first_knowingly(z, y + 1, r, f);
    }
    pull(x, z);
    return res;
  }

  int find_first(int x, int l, int r, int ll, int rr, const function<bool(const node&)> &f) {
    if (ll <= l && r <= rr) {
      if (!f(tree[x])) {
        return -1;
      }
      return find_first_knowingly(x, l, r, f);
    }
    push(x, l, r);
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    int res = -1;
    if (ll <= y) {
      res = find_first(x + 1, l, y, ll, rr, f);
    }
    if (rr > y && res == -1) {
      res = find_first(z, y + 1, r, ll, rr, f);
    }
    pull(x, z);
    return res;
  }

  int find_last_knowingly(int x, int l, int r, const function<bool(const node&)> &f) {
    if (l == r) {
      return l;
    }
    push(x, l, r);
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    int res;
    if (f(tree[z])) {
      res = find_last_knowingly(z, y + 1, r, f);
    } else {
      res = find_last_knowingly(x + 1, l, y, f);
    }
    pull(x, z);
    return res;
  }

  int find_last(int x, int l, int r, int ll, int rr, const function<bool(const node&)> &f) {
    if (ll <= l && r <= rr) {
      if (!f(tree[x])) {
        return -1;
      }
      return find_last_knowingly(x, l, r, f);
    }
    push(x, l, r);
    int y = (l + r) >> 1;
    int z = x + ((y - l + 1) << 1);
    int res = -1;
    if (rr > y) {
      res = find_last(z, y + 1, r, ll, rr, f);
    }
    if (ll <= y && res == -1) {
      res = find_last(x + 1, l, y, ll, rr, f);
    }
    pull(x, z);
    return res;
  }

  segtree(int _n) : n(_n) {
    assert(n > 0);
    tree.resize(2 * n - 1);
    build(0, 0, n - 1);
  }

  template <typename M>
  segtree(const vector<M> &v) {
    n = v.size();
    assert(n > 0);
    tree.resize(2 * n - 1);
    build(0, 0, n - 1, v);
  }

  node get(int ll, int rr) {
    assert(0 <= ll && ll <= rr && rr <= n - 1);
    return get(0, 0, n - 1, ll, rr);
  }

  node get(int p) {
    assert(0 <= p && p <= n - 1);
    return get(0, 0, n - 1, p, p);
  }

  template <typename... M>
  void modify(int ll, int rr, const M&... v) {
    assert(0 <= ll && ll <= rr && rr <= n - 1);
    modify(0, 0, n - 1, ll, rr, v...);
  }

  // find_first and find_last call all FALSE elements
  // to the left (right) of the sought position exactly once

  int find_first(int ll, int rr, const function<bool(const node&)> &f) {
    assert(0 <= ll && ll <= rr && rr <= n - 1);
    return find_first(0, 0, n - 1, ll, rr, f);
  }

  int find_last(int ll, int rr, const function<bool(const node&)> &f) {
    assert(0 <= ll && ll <= rr && rr <= n - 1);
    return find_last(0, 0, n - 1, ll, rr, f);
  }
};

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(n), b(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
    if (b[i] == 1 && 2 * a[i] > k) {
      cout << -1 << '\n';
      return 0;
    }
  }
  vector<int> x(n + 1);
  long long L = 0;
  for (int i = 0; i < n; i++) {
    x[i + 1] = (int) ((x[i] + 2LL * a[i]) % k);
    L += a[i];
  }
  vector<int> x_real = x;
  vector<int> to(n);
  vector<pair<int, int>> xs;
  for (int i = 0; i < n; i++) {
    xs.emplace_back(x[i], i);
    if (b[i] == 1) {
      to[i] = (x[i] + 2 * a[i]) % k;
      xs.emplace_back(to[i], ~i);
    }
  }
  debug(x);
  debug(to);
  sort(xs.begin(), xs.end());
  int t = 0;
  for (int i = 0; i < (int) xs.size(); i++) {
    if (i > 0 && xs[i].first != xs[i - 1].first) {
      ++t;
    }
    int id = xs[i].second;
    if (id >= 0) {
      x[id] = t;
    } else {
      to[~id] = t;
    }
  }
  debug(x);
  debug(to);
  segtree st(t + 1);
  auto Set = [&](int j) {
    if (0 <= j && j < n && b[j] == 1) {
      if (x[j] < to[j]) {
        st.modify(x[j], to[j] - 1, j);
      } else {
        st.modify(x[j], t, j);
        if (to[j] > 0) {
          st.modify(0, to[j] - 1, j);
        }
      }
    }
  };
  vector<bool> start(n);
  for (int i = 0; i < n; i++) {
    Set(i - 2);
    start[i] = (st.get(x[i]).val == -1);
  }
  vector<long long> wait(n);
  long long ans = (long long) 9e18;
  st = segtree(t + 1);
  for (int i = n - 2; i >= 0; i--) {
    Set(i + 1);
    int j = st.get(x[i]).val;
    if (j != -1) {
      wait[i] = wait[j] + ((x_real[i] - x_real[j]) % k + k) % k;
    }
    if (start[i]) {
      debug(i, wait[i]);
      ans = min(ans, wait[i]);
    }
  }
  cout << 2 * L + ans << '\n';
  return 0;
}